Let $f:\mathbb R\to\mathbb R$ be a continuous function and $r\ge0$ a fixed value such that for all $x,y\in\mathbb R$ $$|f(x)+f(y)-f(x+y)|\le r$$ Show there exist $a\in\mathbb R$ and a function $g:\mathbb R\to\mathbb R$ such that for all $x\in\mathbb R$$$|g(x)|\le r,\quad f(x)=g(x)+ax$$
As we might see, if $r=0$, then $f(x)+f(y)=f(x+y)$ and the answer for such a function is $f(x)=cx$ so $$g=0,\quad a=c$$ Satisfy the condition.
But I have difficulties to show this generalization, here's the things I did:
If $|f(x)-ax|\le r$, then: $$-r\le f(x)-ax\le r\,\equiv\, \frac{-r+f(x)}{|x|}\le a\le \frac{r+f(x)}{|x|}$$ So for every $x\neq0$, we have an interval for desired $a$ (also for $x=0$, we can use the inequality and see $f(0)\le r$ so $|f(0)-a\cdot0|\le r$ holds for all $a\in \mathbb R$)
We define sets $$A_x=\{\,a\,|\,a\in[\dfrac{-r+f(x)}{|x|},\dfrac{r+f(x)}{|x|}]\,\}$$ and $A_0=\mathbb R$, I think if we show for all $x,y\in\mathbb R$ $$A_x\cap A_y\neq\emptyset$$ Then problem would be solved, but I was unable to show that.